The vertex of a parabola is at (-4,-3). If one x-intercept is at -11, what is the other x intercept?

The equation of the parabola could be written as y-k = a(x-h)^2, where (h,k) is the vertex.  Thus, y-(-3) = a(x+4)^2, or y+3 = a(x+4)^2.

The coordinates of one x-intercept are (-11,0).  Thus, y+3 = a(x+4)^2 becomes 
0+3 = a(-11+4)^2, so that 3 = a(-7)^2, or 3 = 49a.  Therefore, a = 3/49, and the equation of the parabola becomes

y+3 = (3/49)(x+4)^2.

To find the other x-intercept, let y = 0 and solve the resulting equation for x:

0+3 = (3/49)(x+4)^2, or (49/3)*2 = (x+4)^2

Taking the sqrt of both sides, plus or minus 49/3 = x+4.

plus 49/3 = x+4 results in 37/3 = x, whereas

minus 49/3 = x+4 results in x = -61/3.  Unfortunatelyi, this disagrees with what we are told:  that one x-intercept is x= -11, or (-11,0).


Trying again, using the quadratic equation y=ax^2 + bx + c,
we substitute the coordinates of the points (-4,-3) and (-11,0) and solve for {a, b, c}:

-3 = a(-4)^2 + b(-4) + c, or -3 = 16a - 4b + c

 0 = a(-11)^2 - 11b + c, or 0 = 121a - 11b + c

If the vertex is at (-4,-3), then, because x= -b/(2a) also represents the x-coordinate of the vertex,                       -4 = b / (2a), or  -8a = b, or 
0 = 8a + b

Now we have 3 equations in 3 unknowns:

0  =  8a +  1b
-3 = 16a - 4b + c
0 = 121a - 11b + c

This system of 3 linear equations can be solved in various ways.  I've used matrices, finding that a, b and c are all zero.  This is wrong.


So, let's try again.  Recall that x = -b / (2a) is the axis of symmetry, which in this case is x = -4.  If one zero is at -11, this point is 7 units to the left of x = -4.  The other zero is 7 units to the right of x = -4, that is, at x = 3.

Now we have 3 points on the parabola:  (-11,0), (-4,-3) and (3,0).

This is sufficient info for us to determine {a,b,c} in y=ax^2+bx+c.
One by one we take these 3 points and subst. their coordinates into 
y=ax^2+bx+c, obtaining 3 linear equations:

0=a(-11)^2 + b(-11) + 1c   =>  0 = 121a - 11b + 1c
-3 = a(-4)^2 +b(-4)  + 1c   =>  -3 = 16a   - 4b  +  1c
0 = a(3)^2   +b(3)    + c     =>   0 = 9a     +3b   + 1c

Solving this system using matrices, I obtained a= 3/49, b= 24/49 and c= -99/49.

Then the equation of this parabola, based upon y = ax^2 + bx + c, is

y = (1/49)(3x^2 + 24x - 99)               (answer)

Check:  If x = -11, does y = 0?

(1/49)(3(-11)^2 + 24(-11) - 99 = (1/49)(3(121) - 11(24) - 99
                                                = (1/49)(363 - 264 - 99)  =  (1/49)(0)   YES!

y = (1/49)(3x^2 + 24x - 99)               (answer)